Question

To determine the derivative of the function \(f(x) = \tanh (1 + {e^{2x}})\).

Short answer

The derivative of the function \(f(x) = \tanh \left( {1 + {e^{2x}}} \right)\) is \(2{e^{2x}}\sec {h^2}\left( {1 + {e^{2x}}} \right)\).

Step-by-step solution

Given Information

The equation to be proved is \(\frac{d}{{dx}}\tanh \left( {1 + {e^{2x}}} \right) = 2{e^{2x}}\sec {h^2}\left( {1 + {e^{2x}}} \right)\).

Formula of hyperbolic function

The derivative of\(\tanh x\)is given by\(\frac{d}{{dx}}(\tanh x) = \sec {h^2}x\).

derivation of \(\frac{d}{{dx}}\tanh \left( {1 + {e^{2x}}} \right) = 2{e^{2x}}\sec {h^2}\left( {1 + {e^{2x}}} \right)\)

Evaluate the derivative of\(f(x) = \tanh \left( {1 + {e^{2x}}} \right)\).

\(\begin{array}{c}{f^\prime }(x) = {\left( {\tanh \left( {1 + {e^{2x}}} \right)} \right)^\prime }\\ = \sec {h^2}\left( {1 + {e^{2x}}} \right)\left( {{e^{2x}}} \right)(2)\\ = 2{e^{2x}}\sec {h^2}\left( {1 + {e^{2x}}} \right)\end{array}\)

Therefore, the derivative of the function \(f(x) = \tanh \left( {1 + {e^{2x}}} \right)\) is \(2{e^{2x}}\sec {h^2}\left( {1 + {e^{2x}}} \right)\).